The Mellin Transform

Joubert Oosthuizen

October 2011 1 Introduction

Robert Hjalmar Mellin was a Finnish mathematician who studied under Karl Weier- strass. He is accredited as the developer of the integral transform

Z ∞ M[ f (x); s] = f ∗(s) = f (x)xs−1dx 0 known as the Mellin transform.

We will ﬁrst look at some examples and basic properties of the Mellin transform. Then we proceed to look at the correspondence between the asymptotic expansion of a function and singularities of the transformed function.

Finally we use the Mellin transform in asymptotic analysis for estimating asymptotically harmonic sums.

This project is supported by the National Research Foundation (NRF).

1 2 Basic properties of the Mellin transform

Deﬁnition 2.1 Let f (x) be absolutely integrable on any interval 0 ≤ x ≤ a

Z a i.e. | f (x)|dx < ∞. 0 The largest open strip hα, βi in which the integral converges is called the fundamental strip, where hα, βi is the open strip of complex numbers s = σ + it such that α < σ < β.

Before we can say more about this fundamental strip, we turn our attention to the Laplace transform. Suppose for ﬁnite a that

Z a | f (x)|dx < ∞ 0 and g(t) = O(eαt) as t → ∞ for some real constant α. Then the one-sided Laplace transform Z ∞ L+[g(t); s] = g(t)e−stdt 0 converges absolutely and is analytic in the right-half plane <(s) > α. Similarly suppose Z a |g(−t)|dt < ∞ 0 and g(t) = O(eβt) as t → −∞ for some real constant β.

2 Then the one-sided Laplace transform

Z 0 L−[g(t); s] = g(t)e−stdt −∞ Z ∞ = g(−t)estdt 0 converges absolutely and is analytic in the left half plane Re(s) < β. Under the same conditions on g(t) and if β > α the two-sided Laplace transform

Z ∞ L[g(t); s] = g(t)e−stdt −∞ converges absolutely and is analytic in the vertical strip α < <(s) < β.

Now if we let t = − log x and g(− log x) = f (x) then e−st = es log x = xs.

Hence Z ∞ L[g(t); s] = g(t)e−stdt −∞ Z 0 = − xs−1 f (x)dx ∞ = M[ f (x); s].

So the Mellin transform of f (x) is the two-sided Laplace transform of g(t) where t = − log x and it converges absolutely and is analytic in the vertical strip α < <(s) < β.

Now since g(t) = O(eαt) as t → ∞ we obtain f (x) = O(x−α) as x → 0+.

Also g(t) = O(eβt) as t → −∞ implies f (x) = O(x−β) as x → ∞. Summing up we have proved the following lemma:

Lemma 2.1 The conditions f (x) = O(x−α) as x → 0+ and f (x) = O(x−β) as x → ∞ where α < β guarantee that f ∗(s) exists in the strip hα, βi.

Hence monomials xc, including constants do not have Mellin transforms.

3 Example 2.1

The function f (x) = e−x satisfies e−x = O(x0) as x → 0+ and e−x = O(x−b) as x → ∞ for any b > 0 so that its transform (the gamma function) Z ∞ f ∗(s) = e−x xs−1dx = Γ(s) 0 is defined and analytic on h0, ∞i. Example 2.2 The function f (x) = (ex − 1)−1 satisfies f (x) = O(x−1) as x → 0+ and f (x) = O(x−b) for all b > 0 as x → ∞. Hence f (x) is analytic and defined on h1, ∞i. We find Z ∞ ∗ 1 s−1 f (s) = −x x dx 0 1 − e Z ∞ −x e s−1 = −x x dx 0 (1 − e ) Z ∞ X∞ = e−nx xs−1dx 0 n=1 X∞ Z ∞ = e−nx xs−1dx n=1 0 ∞ X Γ(s) = ns n=1 = Γ(s)ζ(s).

We require that <(s) > 1 for convergence of the Riemann-zeta function and we see that this validates the strip h1, ∞i on which f ∗(s) is deﬁned and analytic. Example 2.3 The function f (x) = (1 + x)−1 is O(x0) as x → 0+ and O(x−1) as x → ∞. Hence a ∗ t guaranteed strip of existence for f (s) is h0, 1i. Set x = 1−t . Then Z 1 t s−1 1 f ∗(s) = (1 − t)−2dt − t 0 1 t 1 + 1−t Z 1 t s−1 = (1 − t)−1dt 0 1 − t Z 1 = ts−1(1 − t)−sdt 0 = β(s, 1 − s) = Γ(s)Γ(1 − s).

4 We can also evaluate the Mellin transform of f (x) using complex analysis. Con- zs−1 sider the branch of the function z+1 deﬁned on the slit plane C\[0, ∞) by rs−1ei(s−1)θ f (z) = ; z = reiθ, 0 < θ < 2π. z + 1 The values on the bottom edge are obtained from those on the top edge by multi- plying by the phase factor e2πi(s−1).

For > 0 small and R > 0 large, we consider the keyhole domain D (see Figure 1 in the appendix) consisting of z in the slit plane C\[0, ∞) satisfying < |z| < R. f (z) has a pole in D, a simple pole at z = −1, with residue

" # zs−1 zs−1 Res , −1 = lim (z + 1) z + 1 z→−1 z + 1 zs = lim z→−1 z = −eπis Hence the residue theorem yields Z f (z)dz = −2πieπis ∂D The integral around δD breaks into the sum of 4 integrals. Z R xs−1 Z zs−1 Z e2πi(s−1) xs−1 Z zs−1 =⇒ −2πieπis = dx + dz + dx + dz 1 + x ΓR z + 1 R 1 + x Γ z + 1

For the integrals over ΓR and Γ we have Z s−1 s−1 z R s−1 dz ≤ 2πR = O R ΓR z + 1 R − 1 Z s−1 s−1 z s dz ≤ 2π = O Γ 1 + z 1 − Since 0 < s < 1 both these integrals vanish as R → ∞ and → 0. Z ∞ xs−1 =⇒ −2πieπis = 1 − e2πi(s−1) dx 0 1 + x

Z ∞ xs−1 −2πieπis =⇒ dx = 2πi(s−1) 0 1 + x 1 − e 2πi = eπis−2πi − e−πis π = sin(πs) π Hence Γ(s)Γ(1 − s) = sin(πs) .

5 Example 2.4 −n x Suppose f (x) = (1 + x) . Setting t = 1+x we obtain Z 1 f ∗(s) = ts−1(1 − t)n−s−1dt 0 = β(s, n − s) Γ(s)Γ(n − s) = . Γ(n) Example 2.5 1 it −it Suppose f (t) = sin t = 2i (e − e ). Consider the region D as in Figure 2 of the appendix.

Since e−tts−1 is analytic in D we have by Cauchy’s theorem: I 0 = e−tts−1dt ∂D Z R Z R Z R = e−uus−1du + i e−(R+iu)(R + iu)s−1du − e−(u+iR)(u + iR)s−1du 0 0 Z R Z π 2 iθ −i e−iu(iu)s−1du − e−e (eiθ)s−1ieiθdθ 0 Letting → 0 and R → ∞ we have:

π Z 2 −eiθ iθ s−1 iθ s e (e ) ie dθ = O( ) → 0 i f Res(s) > 0 0 Z R − − − − e (R+iu)(R + iu)s 1 du = O e RRRe(s 1)R → 0 0 Z R Z R ! − − − − e (u+iR)(u + iR)s 1 du = O RRe(s 1) e udu → 0 i f Res(s) < 1 0 0 Z ∞ Z ∞ =⇒ e−uus−1du = is e−iuus−1du 0 0 Z ∞ Z ∞ −iu s−1 −s −u s−1 −s log i − iπs =⇒ e u du = i e u du = e Γ(s) = e 2 Γ(s) 0 0 Likewise replacing i with −i in the previous equation we obtain Z ∞ iu s−1 iπs e u du = e 2 Γ(s). 0 Combining the two we get Z ∞ s−1 1 − iπs iπs πs (sin t)t dt = e 2 − e 2 = sin Γ(s). 0 2i 2

6 We proceed to look at some functional properties of the Mellin transform.

Theorem 2.1 Let f (x) be a function whose transform admits the fundamental strip hα, βi. Let ρ, µ and v be positive real numbers. Then the following relations hold:

(i) M[ f (µx); s] = µ−s f ∗(s) s ∈ hα, βi     X  X  (ii) M  λ f (µ x); s =  λ µ −s f ∗(s) , µ > 0  k k   k k  k k∈A k∈A (iii) M[xv f (x); s] = f ∗(s + v) s ∈ hα − v, β − vi ! 1 s (iv) M[ f (xρ); s] = f ∗ s ∈ hρα, ρβi ρ ρ

Proof:

(i) M[ f (µx); s] Z ∞ = xs−1 f (µx)dx 0 !s−1 Z ∞ t dt = f (t) 0 µ µ = µ−s f ∗(s).

(ii) Since the Mellin transform is linear this follows immediately from (i).

(iii) M[xv f (x); s] Z ∞ = f (x)xs+v−1dx 0 = f ∗(s + v).

(iv) M[ f (xρ); s] Z ∞ = f (xρ)xs−1dx 0 Z ∞ s −1 dt = f (t)t ρ 0 ρ ! 1 s = f ∗ . ρ ρ

7 −x2 1 s We have for example from Theorem 2.1(iv) M[e ; s] = 2 Γ 2 on h0, ∞i with f (x) = e−x and ρ = 2. Wanting to expand our range of Mellin transforms we ﬁnd by diﬀerentiation under the integral sign: d f ∗(s) = M[ f (x) log x; s] ds For instance the transform of e−x log x is Γ0(s). If we want to transform the derivative of a function integration by parts yields:

Z ∞ Z ∞ h i∞ f 0(x)xs−1dx = f (x)xs−1 − (s − 1) f (x)xs−2dx 0 0 0 h i∞ The term f (x)xs−1 equals 0 since we assume the reason why f ∗(s) exists for 0 s ∈ hα, βi is because lim xρ f (x) = 0 for <(s) > α and lim xρ f (x) = 0 for x→0 x→∞ <(s) < β. Thus M[ f 0(x); s] = −(s − 1) f ∗(s − 1). For instance from Example 2.3 we can derive the Mellin transform of f (x) = log(1 + x) π = M[ f 0(x); s] = −(s − 1)M[log(1 + x); s − 1] sin (πs) π =⇒ M[ f (x); s − 1] = − (s − 1) sin (πs) π π =⇒ M[ f (x); s] = − = s sin (π(s + 1)) s sin (πs) for s ∈ h−1, 0i.

The inversion formula:

The inversion formula for the two-sided Laplace transform of g(t) is given by g(t) = R c i∞ 1 + L st 2πi c−i∞ [g(t); s]e ds for any c inside the region of convergence, provided that g is continuous at t. Setting t = − log x and f (x) = g(− log x) we obtain the inversion formula for the Mellin transform: 1 Z c+i∞ f (x) = g(− log x) = M[ f (x); s]x−sds 2πi c−i∞ We will prove the inversion formula for the Laplace transform, from which the inversion formula of the Mellin transform follows. Before we can do this we need to prove the following lemma (Riemann-Lebesgue):

Lemma 2.2 R b iωt → → ∞ I(ω) = a f (t)e dt 0 as ω if f (t) is continuous on [a, b].

8 Proof:

π Z a+ ω Z b I(ω) = f (t)eiωtdt + f (t)eiωtdt π a a+ ω Z a+ π Z b− π ω ω π = f (t)eiωtdt − f t + eiωtdt a a ω and π Z b− ω Z b I(ω) = f (t)eiωtdt + f (t)eiωtdt π a b− ω Z a+ π Z b Z b− π 1 ω 1 1 ω π =⇒ I(ω) = f (t)eiωtdt+ f (t)eiωtdt+ f (t) − f t + eiωtdt 2 2 π 2 ω a b− ω a In the ﬁrst two integrals f (t)eiωt is bounded, since it is a continuous function on a closed interval. Since the length of these integrals goes to 0 as ω → ∞ they both π vanish. Similarly for the last integral we ﬁnd f (t) − f t + ω → 0 as ω → ∞ and since the length of this integral is bounded it also vanishes so that lim I(ω) = 0 ω→∞

Note that the proof also holds for an inﬁnite interval if

Z ∞ | f (t)|dt < ∞ : 0 Choose T large enough so that for all > 0

Z ∞ | f (t)|dt ≤ T

Z ∞ Z T Z ∞ =⇒ f (t)eiωtdt = f (t)eiωtdt + f (t)eiωtdt 0 0 T Z ∞ =⇒ lim f (t)eiωtdt = 0 ω→∞ 0 since is arbitrary.

Dirichlet conditions and Dirichlet integrals:

f (t) satisfies the Dirichlet conditions if it only has a finite number of discontinuities with finite jumps and a finite number of minima/maxima. (i.e. f is piecewise monotone) The second mean value theorem for integrals states: If f (t) is monotone [a, b], then for any continuous g(t)

Z b Z c Z b f (t)g(t)dt = f (a) g(t)dt + f (b) g(t)dt a a c

9 for some c ∈ [a, b]. Assume f (t) satisﬁes the Dirichlet conditions. We would like to determine

Z b sin ωt Z b eiωt lim dt = lim Im dt ω→∞ 0 t ω→∞ 0 t Choose c such that f (t) is continuous and monotone on [0, c].

Z b sin ωt Z c sin ωt Z b sin ωt f (t) dt = f (t) dt + f (t) dt 0 t 0 t c t Z b sin ωt lim dt = 0 ω→∞ c t by the Riemann-Lebesgue lemma. Apply the mean value theorem:

Z c sin ωt Z h sin ωt Z c sin ωt f (t) dt = f (0+) dt + f (c) dt 0 t 0 t h t Z c sin ωt Z c sin ωt = f (0+) dt + f (c) − f (0+) dt 0 t h t

Z c sin ωt Z ωc ω sin x dx dt = 0 t 0 x ω Z ωc sin x = dx 0 x Z ω sin x = lim dx ω→∞ 0 x

R ∞ sin x 0 x dx can not be evaluated by elementary analytic methods. To evaluate it we use complex analysis and the following lemma (Jordan’s lemma): Lemma 2.3 If Γ is the semicircular contour z(θ) = Reiθ where 0 ≤ θ ≤ π, in the R R upper half-plane, then |eiz||dz| < π. ΓR Z R sin x π Now we show that lim dx = . R→∞ 0 x 2 Z R sin x Since the integrand is an even function it is equivalent to lim dx = π R→∞ −R x To evaluate the limit we consider f (z) = eiz which only has one pole, a simple pole z at = 0. " # eiz Res , 0 = 1 z Consider Figure 3 in the appendix:

The integrand around ∂D breaks into 4:

10 Z Z − Z Z R Z ! f (z)dz = + + + f (z)dz δD −R C ΓR Since f is analytic on D, Cauchy’s theorem gives: Z 0 = f (z)dz (∗) ∂D

The integrand C is handled by the fractional residue theorem with angle −π: " # Z eiz eiz lim dz = −πiRes , 0 = −πi →0 C z z Taking the limit as → 0 in (∗) we obtain:

Z R eix Z eiz 0 = dx − πi + dz −R x ΓR z Taking imaginary parts gives:

Z R sin x Z eiz π = dx + Im dz −R x ΓR z

Now the integral over ΓR is handled by Jordan’s lemma: Z Z eiz eiz π dz ≤ |dz| < → 0 ΓR z ΓR z R Z R sin x as R → ∞. Thus we have lim dx = π. Hence R→∞ −R x Z R sin x π lim dx = R→∞ 0 x 2 Returning to where we were just before Jordan’s lemma we have for the other part: Z c sin ωt f (c) − f (0+) dt → 0 h t if ω → ∞, since f (c) − f (0+) can be made arbitrarily close to 0 by the choice of c Z c sin ωt Z b sin ωt π and since dt is bounded. It follows that lim f (t) dt = f (0+). h t ω→∞ 0 t 2 Z 0 sin ωt π Likewise lim f (t) dt = f (0−). By the Riemann-Lebesgue lemma we ω→∞ −b t 2 have Z ∞ sin ωt lim f (t) dt = 0. ω→∞ b t Hence: Z ∞ sin ωt π lim f (t) dt = f (0+). ω→∞ 0 t 2

11 Z 0 sin ωt π lim f (t) dt = f (0−) ω→∞ −∞ t 2 So that Z ∞ sin ωt π lim f (t) dt = f (0+) + f (0−) ω→∞ −∞ t 2 Now we can ﬁnally prove the inversion formula of the one-sided Laplace transform. Assume f is continuous at t. Since the integral of the Laplace transform is uniformly convergent we can switch integrals in the following expression: ! 1 Z c+iR 1 Z c+iR Z ∞ F(s)estds = f (u)e−sudu estds 2πi c−iR 2πi c−iR 0 1 Z ∞ Z c+iR = f (u) es(t−u)dsdu 2πi 0 c−iR 1 Z ∞ Z R = f (u) e(c+ir)(t−u)idrdu , s = c + ir 2πi 0 −R 1 Z ∞ Z R = f (u)ec(t−u) eir(t−u)drdu 2π 0 −R

Z R e(t−u)iR − e−(t−u)iR eir(t−u)dr = −R i(t − u) 2 sin ((t − u)R) = t − u Now 1 Z c+iR 1 Z ∞ sin ((u − t)R) F(s)estds = f (u)ec(t−u) du 2πi c−iR π 0 u − t 1 Z ∞ sin (vR) = f (v + t)e−cv dv π −t v Since f is continuous if we let R → ∞ the integral converges to π f (t+) + f (t−) = π f (t) 2 Z c+i∞ 1 st So f (t) = 2πi F(s)e ds if f satisﬁes the Dirichlet conditions and is contin- c−i∞ uous at t. By the change of variable previously mentioned we obtain the inversion formula for the Mellin transform from the two-sided Laplace transform which is a sum of two one-sided Laplace transforms. Theorem 2.2 (Inversion formula) Let f (x) be absolutely integrable with fundamental strip hα, βi. If f satisﬁes the Dirichlet conditions and is continuous at x then for any c in the interval (α, β) 1 Z c+i∞ f (x) = f ∗(s)x−sds . 2πi c−i∞

12 3 The fundamental correspondence

There is a correspondence between the asymptotic expansion of a function at 0 and ∞ and poles of its Mellin transform in a left and respectively right half-plane. The correspondence fares both ways and forms the basis of an asymptotic process: For estimating asymptotically a function f (x), determine its Mellin transform and translate back its singularities into asymptotic terms in the expansion of f (x).

3.1 Singularities

If φ(s) is meromorphic at s = s0 it admits a Laurent expansion near s0: X k φ(s) = ck(s − s0) . k≥−r

The function φ(s) has a pole of order r if r > 0 and c−r , 0.

Deﬁnition 3.1 A singular element of φ(s) at s0 is an initial sum of the Laurent expansion truncated at terms of order O(1) or smaller.

Example 3.1 1 Consider the function φ(s) = , which has two poles in the complex plane, s2(s + 1) a double pole at s0 = 0 and a simple pole at s0 = −1. 1 1 1 = s2(s + 1) s + 1 (1 − (s + 1))2 ∞ ! 1 X −2 = [−(s + 1)]n s + 1 n n=0 1 = + 2 + 3(s + 1) + ... s + 1

1 1 1 = s2(s + 1) s2 1 − (−s) 1 1 = − + 1 − s + ... s2 s

Singular elements at s0 = 0 are: " # " # " # 1 1 1 1 1 1 − , − + 1 , − + 1 − s , ... s2 s s2 s s2 s We can truncate the Laurent expansion wherever we want as long as we include all the terms with negative degree of s. Similarly we have the following singular elements for s = −1: 0 " # " # 1 1 , + 2 , ... s + 1 s + 1

13 Deﬁnition 3.2 Let φ(s) be meromorphic in Ω with P including all the poles of φ(s) in Ω. A singular expansion of φ(s) in Ω is a formal sum of singular elements of φ(s) at all points of P. When E is a singular expansion of φ(s) in Ω, we write φ(s) E (s ∈ Ω).

Example 3.2

For instance " # " # " # 1 1 1 1 1 + − + 2 2 s (s + 1) s + 1 s=−1 s s s=0 2 s=1 This expansion is a concise way of combining information contained in the Laurent expansions of the function φ(s) ≡ s−2(s+1)−1 at the three points of P = { −1, 0, 1} . Note that there is no pole at s0 = 1, but we can include an extra singular element if we want. P is to include all poles of the function but may also include other points. Singular expansion is a formal sum which we do not want to evaluate. It only shows us which poles a function have.

Example 3.3

Recall the gamma function

Z ∞ Γ(s) = e−x xs−1dx 0 for s ∈ h0, ∞i. The well-known functional equation Γ(s + 1) = sΓ(s) follows if we integrate by parts. This allows us to extend Γ(s) to a meromorphic function in the whole of C. We have from the functional equation: Γ(s + m + 1) Γ(s) = s(s + 1)(s + 2)...(s + m)

(−1)m 1 Thus Γ(s) has poles at the points s = −m , m ∈ N ∪ { 0 } and Γ(s) ∼ m! s+m as s → −m .This means the right side is a singular element of Γ(s) at s = −m. Hence we obtain the following singular expansion in the whole of C

∞ X (−1)k 1 Γ(s) . k! s + k k=0

3.2 Direct mapping e−x has a Taylor expansion at x = 0 :

∞ X (−1)k e−x = xk k! k=0

14 Comparing this with the singular expansion of Γ(s) we see there is a striking coin- k 1 cidence of coeﬃcients expressed by the rule x 7→ s+k . We will see that this is a general phenomenon. Some information about the connection between the asymptotic expansion of our initial function and properties of its Mellin transform are summed up in the following table:

f (x) f ∗(s) Order at 0: O(x−α) Leftmost boundary of f.s. at Re(s) = α Order at ∞: O(x−β) Rightmost boundary of f.s. at <(s) = β Expansion till O(xγ) at 0 Meromorphic continuation till <(s) = −γ Expansion till O(xδ) at ∞ Meromorphic continuation till <(s) = −δ

The following table will follow from the Direct mapping theorem:

f (x) f ∗(s) (−1)kk! Term xa(log x)k at 0 Pole with singular element (s + a)k+1 (−1)kk! Term xa(log x)k at ∞ Pole with singular element − (s + a)k+1

Visualising these two tables:

Theorem 3.1

Let f (x) have Mellin transform f ∗(s) with non-empty fundamental strip hα, βi. (i) Assume that f (x) admits as x → 0+ a ﬁnite asymptotic expansion of the form X a k γ f (x) = ca,k x (log x) + O(x ) (a,k)∈ A where −γ < −a ≤ α and the k are nonnegative. Then f ∗(s) is continuable to a meromorphic function in the strip h−γ, βi where it admits the singular expansion k ∗ X (−1) k! f (s) ca,k , s ∈ h−γ, βi. (s + a)k+1 (a,k)∈ A

15 (ii) Similarly assume that f (x) admits as x → ∞ a ﬁnite asymptotic expansion of the form X a k γ f (x) = ca,k x (log x) + O(x ) (a,k)∈ A where now β ≤ −a ≤ −γ. Then f ∗(s) is continuable to a meromorphic function in the strip hα, −γi where k ∗ X (−1) k! f (s) − ca,k , s ∈ hα, −γi. (s + a)k+1 (a,k)∈ A Thus terms in the asymptotic expansion of f (x) at 0 induce poles of f ∗(s) in a strip to the left of the fundamental strip. Terms in the expansion at ∞ induce poles in a strip to the right.

Proof: h 1 i Since M f x ; s = M ( f (x)) ; −s we only need to treat case (i) corresponding → + to x 0 . X a k γ By assumption g(x) = f (x) − ca,k x (logx) satisfies g(x) = O(x ). For (a,k)∈ A s ∈ hα, βi we have: Z 1 Z ∞ f ∗(s) = f (x)xs−1dx + f (x)xs−1dx 0 1 Z Z   Z ∞ 1 1  X  = g(x)xs−1dx +  c xa(logx)k xs−1dx + f (x)xs−1dx  a,k  0 0 (a,k)∈ A 1 In the last line the first integral defines an analytic function of s in the strip h−γ, βi since g(x) = O(xγ) as x → 0 , the third integral is analytic in h−∞, βi so that the sum of these two are analytic in h−γ, βi. Finally   Z 1  X  X k  a k s−1 (−1) k!  ca,k x (logx)  x dx = ca,k   (s + a)k+1 0 (a,k)∈ A (a,k)∈ A This result follows since Z 1 k Z 1 (logx)k xs+a−1dx = − (logx)k−1 xs+a−1dx 0 s + a 0 from integration by parts. Integrating by parts k times yields the above result. Now X (−1)kk! ca,k is meromorphic in all of C and provides the singular expan- (s + a)k+1 (a,k)∈ A sion of f ∗(s) in the extended strip. This completes the proof of the Direct mapping theorem.

Note that in the case where there exists a complete expansion of f (x) at 0 ( or ∞ ), the transform f ∗(s) becomes meromorphic in a complete left ( or right ) half-plane. This situation always occurs for functions that are analytic at 0 ( or ∞).

16 Example 3.4

The Riemann zeta function plays a pivotal role in analytic number theory and has applications in physics, probability theory and applied statistics. Using singular expansion and the Direct mapping theorem we can easily derive some of its properties. e−x ∗ Recall Example 2.2 where we showed if f (x) = 1−e−x then f (s) = Γ(s)ζ(s) with fundamental strip h1, ∞i. f (x) is exponentially small at inﬁnity and it admits a complete expansion near x = 0:

∞ e−x X xk = B 1 − e−x k+1 (k + 1)! k=−1

1 1 which deﬁnes the Bernoulli numbers Bk: B0 = 1, B1 = − 2 , B2 = 6 , ...

Thus f ∗(s) = Γ(s)ζ(s) is meromorphic on the whole of C with singular expan- ∞ X B 1 sion Γ(s)ζ(s) k+1 . If we compare it with the singular expansion of (k + 1)! s + k k=−1 ∞ X (−1)k 1 the gamma function Γ(s) we can extract the singular expansion k! s + k k=0 of the zeta function " # ∞ " # 1 X B j+1 ζ(s) + (−1) j s − 1 j + 1 s=1 j=0 s=− j

Hence the zeta function is meromorphic in the whole of C with only a simple pole at s = 1. It follows from the singular expansion that B ζ(−m) = (−1)m m+1 , m ∈ N ∪ { 0 }. m + 1

1 Hence ζ(0) = − 2 . Since B2k+1 = 0 for k ∈ N it follows that ζ(−2m) = 0 for all m ∈ N. These zeros are called the trivial zeros of the zeta function. The nontrivial zeros of the zeta function lie in the strip { 0 ≤ <(s) ≤ 1 } which is called the critical strip.

The most famous unsolved problem in complex analysis is the Riemann hypothesis 1 which states that the nontrivial zeros of the zeta function lie on the line {<(s) = 2 }. It is also one of the Clay Mathematics Institute Millennium Prize Problems. Computer calculations have shown that the ﬁrst 10 trillion zeros lie on the critical line. Whether the Riemann hypothesis is true remains a famous unresolved problem.

17 Example 3.5

−1 ∗ π Recall from Example 2.3 the Mellin pair f (x) = (1 + x) , f (s) = sin πs with fundamental strip h0, 1i. As x → 0+ :

∞ 1 X = (−1)n xn 1 + x n=0 and as x → ∞ : ∞ 1 x−1 X = = (−1)n−1 x−n 1 + x 1 + x−1 n=1 These expansions translate into:

∞ X (−1)n f ∗(s) s ∈ h−∞, 1i s + n n=0 which is the continuation of the transformed function to the left of the fundamental strip and ∞ X (−1)n−1 f ∗(s) − s ∈ h0, ∞i s − n n=1 which is the continuation of the transformed function to the right of the fundamental strip. We can combine the two singular expansions into one, giving us a singular expansion in the whole of C : π X (−1)n f ∗(s) = , s ∈ C. sin πs s + n n ∈ Z

3.3 Converse mapping: Under a set of mild conditions, a converse to the Direct mapping theorem also holds: The singularities of a Mellin transform which is small enough towards ±i∞ encode the asymptotic properties of the original function.

Theorem 3.2

Let f (x) be continuous with Mellin transform f ∗(s) having a non-empty fundamental strip hα, βi. (i) Assume that f ∗(s) admits a meromorphic continuation to the strip hγ, βi for some γ < α with a ﬁnite number of poles there, and is analytic on <(s) = γ. As- sume also that there exists a real number η ∈ (α, β) such that f ∗(s) = O |s|−r with r > 1 when |s| → ∞ in γ ≤ <(s) ≤ η. If f ∗(s) admits the singular expansion for s ∈ hγ, αi , ∗ X 1 f (s) da,k (s − a)k (a,k)∈ A

18 then an asymptotic expansion of f (x) at 0 is ! X (−1)k−1 f (x) = d x−a(log x)k−1 + O(x−γ). a,k (k − 1)! (a,k)∈ A (ii) Similarly assume that f ∗(s) admits a meromorphic continuation to hα, γi for some γ > β and is analytic on <(s) = γ. Assume again that f ∗(s) = O |s|−r with r > 1 when |s| → ∞ in hη, γi for some η ∈ (α, β). If

∗ X 1 f (s) da,k (s − a)k (a,k)∈ A for s ∈ (η, γ) , then the asymptotic expansion of f (x) at ∞ is ! X (−1)k−1 f (x) = − d x−a(log x)k−1 + O(x−γ). a,k (k − 1)! (a,k)∈ A Proof: As in the direct mapping theorem, it suﬃces to consider case (i) corresponding to continuation to the left. Consider the integral 1 Z I(T) = f ∗(s)x−sds 2πi ∂D where D denotes the rectangular contour deﬁned by the segments [η − iT, η + iT] , [η + iT, γ + iT] , [γ + iT, γ − iT] , [γ − iT, η − iT]. By the residue theorem I(T) is equal to the sum of residues, which are: ! X x−s R = da,k Res (s − a)k (a,k)∈ A s=a ! X (−1)k−1 = d x−a(log x)k−1 a,k (k − 1)! (a,k)∈ A Let T → ∞. The integrals along the two horizontal segments vanish, since they are O(T −r) and tend to 0 as T → ∞. The integral along the vertical line <(s) = η lies inside the fundamental strip and tends to the inverse Mellin integral f (x) by the inversion theorem since f (x) is continuous. For the integral along the vertical line <(s) = γ we have: Z γ+i∞ Z γ+i∞ 1 ∗ −s 1 ∗ −<(s) f (s)x ds ≤ | f (s)|x |ds| 2πi γ−i∞ 2π γ−i∞ Z ∞ x−γ = O(1) r dt 0 (1 + t) = O(x−γ) Hence for T → ∞ we have R = f (x) + O(x−γ). So that f (x) = R + O(x−γ). This completes the proof of the Converse mapping theorem. The following table sums up aspects of the Converse mapping theorem:

19 f ∗(s) f (x) Pole at a Term in asymptotic expansion ≈ x−a left of fundamental strip expansion at 0 right of fundamental strip expansion at ∞

Simple pole 1 −a left: s−a x at 0 1 −a right: s−a −x at ∞

Multiple pole Logarithmic factor 1 (−1)k −a k left: (s−a)k+1 k! x (log x) at 0 k 1 − (−1) −a k ∞ right: (s−a)k+1 k! x (log x) at

Example 3.6

−v Γ(s)Γ(v−s) In example 2.4 we showed that (1 + x) has Mellin transform Γ(v) . Now Γ(s)Γ(v−s) consider φ(s) = Γ(v) that is analytic in the strip h0, vi. We use the singular expansion of the gamma function to obtain the singular expansion to the left of <(s) = 0 ( i.e. for s ∈ h−∞, vi ): (−1)k Γ(v + k) 1 φ(s) . k Γ(v) s + k Since the gamma function decreases along vertical lines it follows that φ(s) decreases along vertical lines, so that the conditions of the converse mapping theorem is satisﬁed. φ(s) is the Mellin transform of some function f (x) , which can be obtained by the inversion formula: Z v +i∞ 1 2 f (x) = φ(s)x−sds 2πi v 2 −i∞ Singularities of φ(s) in h−∞, vi encode asymptotics for f (x) at 0:

∞ X (−1)k Γ(v + k) f (x) = xk k! Γ(v) k=0

∞ ! X −v Furthermore (1 + x)−v = xk k k=0 ∞ X (−1)k = v(v + 1)(v + 2)...(v + k − 1)xk k! k=0 ∞ X (−1)k Γ(v + k) = xk k! Γ(v) k=0

20 1 Γ(s)Γ(v − s) Hence we again found the Mellin pair f (x) = , f ∗(s) = , this (1 + x)v Γ(v) time using the Inversion formula and the Converse mapping theorem.

21 4 Application to Harmonic Sums

In this section we look at an application of the Mellin transform to harmonic sums using the functional properties of the Mellin transform and the fundamental correspondence. X Deﬁnition 4.1 A sum of the form G(x) = λkg(µk x) is called a harmonic sum k with base function g(x) , frequencies µk and amplitudes λk.

X −s The series Λ(s) = λk(µk) is called a Dirichlet series. Recall the following k useful property of the Mellin transform: M[g(µx), s] = µ−sg∗(s). X ∗ ∗ Hence if G(x) = λkg(µk x) then G (s) = Λ(s)g (s). k So we have the following proposition: X Proposition 4.1 The Mellin transform of the harmonic sum G(x) = λkg(µk x) k is deﬁned in the intersection of the fundamental strip of g(x) and the domain of X −s absolute convergence of the Dirichlet series Λ(s) = λk(µk) ( which is of the k ∗ ∗ form Re(s) > σa for some real σa ) and is given by G (s) = Λ(s)g (s).

This is illustrated in the following picture:

Example 4.1

Consider the function ∞ " # X 1 1 h(x) = − k k + x k=1 ∞ X 1 x = k k 1 + x k=1 k

22 1 1 x This is a harmonic sum with µk = k , λk = k and base function g(x) = 1+x . 1 1 Note that h(n) = 1 + 2 + ... + n = Hn , the harmonic number. X λ Λ(s) = k µ s k k X∞ = ks−1 k=1 = ζ(1 − s) π π Transform of the base function is g∗(s) = = − with fundamental sin π(s + 1) sin πs strip h−1, 0i = h0 − 1, 1 − 1i , which follows when applying theorem 2.1(iii) to example 2.3. Now π h∗(s) = Λ(s)g∗(s) = −ζ(1 − s) , s ∈ h−1, 0i. sin πs 1 We showed that ζ(s) ∼ s−1 is the beginning of the Laurent expansion of the zeta function. It is well known that the coeﬃcient of zero degree in the Laurent expansion is γ ≈ 0.57721566 , the Euler-Mascheroni constant. This can be found in 1 [5]. Hence ζ(s) = s−1 + γ + ... as s → 1. Hence we have the following singular expansion of h∗(s): " # ∞ 1 γ X ζ(1 − k) h∗(s) − − (−1)k , s ∈ h−1, ∞i. s2 s s − k k=1

m Bm+1 Remembering that ζ(−m) = (−1) m+1 for m ∈ N ∪ {0} , we use the converse mapping theorem and obtain: 1 X (−1)kB 1 1 1 1 H ∼ log n + γ + − k = log n + γ + − + − ... n 2n k nk 2n 12n2 120n4 k≥2 Example 4.2 It follows from the product decomposition of the Gamma function that ∞ X x x l(x) = log Γ(x + 1) + γx = − log 1 + . n n n=1 1 This is a harmonic sum with amplitudes λn = 1 , frequencies µn = n and base function g(x) = x − log(1 + x). The Dirichlet series is ∞ X −s Λ(s) = λn(µn) n=1 X∞ = ns n=1 = ζ(−s).

23 ∗ π Remember that the Mellin transform of f (x) = − log (1 + x) is f (s) = − s sin πs for s ∈ h−1, 0i. Notice that g(x) = x[1 − x−1 log(1 + x)]. 1 − x−1 log(1 + x) is O(x) as x → 0+ and O x0 as x → ∞. Hence 1 − x−1 log(1 + x) is f ∗(s − 1) for ∗ ∗ π s ∈ h−1, 0i. This means g (s) = f (s) = − s sin πs for s ∈ h−2, −1i. Since ζ(−s) ∗ π converges absolutely in this fundamental strip we have: l (s) = −ζ(−s) s sin πs for s ∈ h−2, −1i. To obtain an asymptotic expansion at ∞ we must ﬁnd a meromorphic continuation to the right of the fundamental strip. The Laurent expansion of the zeta function 1 1 2 at 0 is ζ(s) = − 2 − 2 log 2π(s) + O(s ). Taking note that there are double poles at s = −1, s = 0 and simple poles at the positive integers, we obtain the following singular expansion: √ " #   ∞ n−1 ∗ 1 1 − γ  1 log 2π X (−1) ζ(−n) l (s) + +  −  + , s ∈ h−2, ∞i. (s + 1)2 s + 1 2s2 s  n(s + n) n=1 The Converse mapping theorem yields as x → ∞:

" # ∞ 1 √ X ζ(−n)(−1)n−1 l(x) ∼ x log x + γx − x + log x + log 2π + . 2 n(−xn) n=1 Now Γ(x + 1) = x! and

∞ ∞ X ζ(−n)(−1)n−1 X B (−1)n−1 = n+1 n(−xn) (n + 1)(n)xn n=1 n=1 ∞ X B = 2n . (2n)(2n − 1)x2n−1 n=1

The last line follows since we can replace n by 2n − 1 because B2k+1 = 0 for all k ∈ N. Putting this all together we obtain Stirling’s formula:

∞ √ X B log (x!) ∼ log xxe−x 2πx + 2n . (2n)(2n − 1)x2n−1 n=1

24 5 Appendix

D ΓR

−R R

Figure 1: Domain D.

iR D

Figure 2: Domain D.

25 D

ΓR

C 0 −R − R

Figure 3: Domain D.

References

[1] P.Flajolet, X.Gourdon and P.Dumas, Mellin transforms and asymptotics: Har- monic sums, Theoretical Computer Science 144 (1995) 3-58.

[2] T.W. Gamelin, Complex Analysis (Springer, New York, 2001)

[3] J.Gajjar, Mathematical Methods (2008). Available as maths.dept.shef.ac.uk/magic/course ﬁles/41/notes week8a.pdf

[4] B.Davies, Integral Transforms and Their Applications (Springer, Berlin 1978)

[5] T.M. Apostol, Introduction to Analytic Number Theory (Springer, Berlin, 1976)

[6] I.Posov, Mellin transforms and asymptotics: Harmonic sums (2004). Available as www14.informatik.tu-muenchen.de/konferenzen/Jass04/courses/1/presentations/11